(a) Solve the equation : \(\sqrt{4x - 3} - \sqrt{2x - 5} = 2\).
(b) Find the finite area enclosed by the curve \(y^{2} = 4x\) and the line \(y + x = 0\).
(a) \(\sqrt{4x - 3} - \sqrt{2x - 5} = 2\)
Squaring both sides, we have
\(4x - 3 - 2\sqrt{8x^{2} - 26x + 15} + 2x - 5 = 4\)
\(6x - 8 - 4 = 2\sqrt{8x^{2} - 26x + 15}\)
\(6x - 12 = 2\sqrt{8x^{2} - 26x + 15}\)
\(3x - 6 = \sqrt{8x^{2} - 26x + 15}\)
Squaring both sides, we have
\(9x^{2} - 36x + 36 = 8x^{2} - 26x + 15\)
\(9x^{2} - 8x^{2} - 36x + 26x + 36 - 15 = 0\)
\(x^{2} - 10x + 21 = 0\)
Factorising, we have
\((x - 7)(x - 3) = 0\)
\(\text{x = 7 or 3}\).
(b) 
We integrate with respect to y.
\(y^{2} = 4x \implies x = \frac{y^{2}}{4}\)
\(x + y = 0 \implies x = -y\)
To get the boundary, equate the two values of x;
\(\frac{y^{2}}{4} = - y \implies y^{2} = -4y\)
\(y^{2} + 4y = 0 \implies y(y + 4) = 0\)
\(y = 0; y = -4\)
\(\int_{-4}^{0} (\frac{y^{2}}{4}) \mathrm {d} y = [\frac{y^{3}}{12}]|_{-4}^{0}\)
\(0 - \frac{-4^{3}}{12} = 0 - \frac{-64}{12} = \frac{64}{12} sq. units\)
\(\int_{-4}^{0} - y \mathrm {d} y = [-\frac{y^{2}}{2}]|_{-4}^{0}\)
\(0 - \frac{-(-4^{2})}{2} = \frac{16}{2} = 8 sq. units\)
Area enclosed = \(8 - \frac{64}{12} = \frac{32}{12} = 2\frac{2}{3} sq. units\)
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